java - 如何比较表示为 long 和 int 的小数？

Question

有没有办法可以比较(>、<、>=、<=、!=、==)表示为 long 和 int 的小数？

如果数字是 3214.21 那么它将在这样的类中表示

long units = 321421;
int precision = 2;
// to get the original number I would do units * 10^precision

我希望能够执行类似于 BigDecimal 的compareTo() 方法的操作。因此大于返回 1，等于返回 0，小于返回 -1。

我目前所做的事情在某些情况下不起作用。下面概述了使其以这种方式运行的代码。该方法或多或少是一个概念证明。

public int compareTo(Money other) {
    if (precision == other.getPrecision()) { // fast check if precision is the same
        if (units > other.getUnits()) return 1; // we forgot to inverse/flip here. will be an issue for non-decimal
        else if (units < other.getUnits()) return -1;
        else return 0; // least likely
    }

    int intX = (int) (units / (Math.pow(10, precision))); // converted units whole numbers to int
    int fractionX = (int) (units % (Math.pow(10, precision))); // converts the decimal as an int

    int intY = (int) (other.getUnits() / (Math.pow(10, other.getPrecision()))); // converted units whole numbers to int
    int fractionY = (int) (other.getUnits() % (Math.pow(10, other.getPrecision()))); // converts the decimal as an int

    System.out.println("Test: i " + intX + "| f " + fractionX + "| u " + units + "| p " + precision);
    System.out.println("Test2: i " + intY + "| f " + fractionY + "| u " + other.getUnits() + "| p" + other
    .getPrecision
    ());

    if (intX > intY) return 1;
    else if (intX < intY) return -1;
    else {
        if (fractionX > fractionY) return 1; // this is where the logic fails
        if (fractionX < fractionY) return -1;
        else return 0;
    }
}

这是我的测试以及输出

System.out.println(MoneyFactory.fromString("0.3").compareTo(MoneyFactory.fromString("0.29")));

System.out.println(MoneyFactory.fromString("13").compareTo(MoneyFactory.fromString("0.31456789")));

System.out.println(MoneyFactory.fromString("0.2999").compareTo(MoneyFactory.fromString
("0.3")));

输出

Test: i 0| f 3| u 3| p 1
Test2: i 0| f 29| u 29| p2
-1
Test: i 13| f 0| u 13| p 0
Test2: i 0| f 31456789| u 31456789| p8
1
Test: i 0| f 2999| u 2999| p 4
Test2: i 0| f 3| u 3| p1
1

Answer 1

最简单的解决方案是将一个数字转换为通用精度级别，然后比较这些数字。如果它们相同，则使用“具有更大精度的数字”逻辑(伪代码):

return (number1 == number2) ? [number with bigger precision logic] : number1 - number2

在Java代码中

class Money {
   long units;
   int precision;

   public Money (long un, int prec)  {
     units =  un;
     precision = prec;
   }

   public int compareTo(Money other) {
      int preResult = this.precision - other.precision;
      long first =  (preResult > 0) ?  ((long)(this.units / Math.pow(10, preResult))) : this.units;
      long second = (preResult < 0) ?  ((long)(other.units * Math.pow(10, preResult))) : other.units;
      return (first == second) ? preResult : Long.compare(first, second);
   }

   public static void test() {
      Money first  = new Money(2345L, 4);
      Money second = new Money(234567L, 6);
      System.out.println(first.compareTo(second));
   }

}

编辑:代码中有错误。将两个长期检查中的 1 更改为 0 可修复此问题